Question

Derive an expression for Ostwald’s dilution law.

Answer 1

Ostwald’s dilution law: It relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).

Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as,

\[\ce{CH3COOH ⇌ CH3COO^- + H^+}\]

The dissociation constant of acetic acid is,

K_a = `(["H"^+]["CH"_3"COO"^-])/(["CH"_3"COOH"])`

	CH3COOH	H+	CH3COO−
Initial number of moles	1	-	-
Degree of dissociation of CH3COOH	α	-	-
Number of moles at equilibrium	1 − α	α	α
Equilibrium concentration	(1 − α) C	α C	α C

Substituting the equilibrium concentration in the equation

K_a = `((α  "C")(α  "C"))/((1 - α)  "C")`

K_a = `(α^2  "C"^2)/((1 - α)  "C")`

K_a = `(α^2  "C")/((1 - α))` ........................(1)

We know that weak acid dissociates only to a very small extent compared to one, a is so small.

equation (1) becomes,

K_a = α² C

α² = `"K"_"a"/"C"`

α = `sqrt(("K"_"a")/"C")` ...............(2)

Similarly, for a weak base,

K_b = α² C

α = `sqrt(("K"_"b")/"C")` ...............(3)

The concentration of H can be calculated using the K_a value as below,

[H⁺] = α C

α = `(["H"^+])/"C"`

Substituting a value in equation (2),

`(["H"^+])/"C" = sqrt("K"_"a"/"C")`

[H⁺] = `sqrt("K"_"a"/"C")."C"`

[H⁺] = `sqrt(("K"_"a". "C"^2)/"C")`

[H⁺] = `sqrt("K"_"a"."C")`

For weak base

[OH⁻] = `sqrt("K"_"b"."C")`